In a `DeltaABC,` sides a,b,c are inAP and `(2)/(1!9!)+(2)/(3!7!)+(1)/(5!5!)=(8^(a))/((2b)!),` then the maximum value of tan A tan B is equal to

Options

A. `1/2`
B. `1/3`
C. `1/4`
D. `1/5`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the given information
We are given that in triangle ABC, the sides a, b, and c are in Arithmetic Progression (AP). This means that:
\[ 2b = a + c \]

### Step 2: Analyze the equation
We need to analyze the equation:
\[ \frac{2}{1!9!} + \frac{2}{3!7!} + \frac{1}{5!5!} = \frac{8^a}{(2b)!} \]

### Step 3: Simplify the left-hand side (LHS)
The LHS can be rewritten as:
\[ \frac{1}{1!9!} + \frac{1}{3!7!} + \frac{1}{5!5!} \]
This can be recognized as a combinatorial expression. We can express it in terms of binomial coefficients.

### Step 4: Use the identity for binomial coefficients
Using the identity for binomial coefficients, we can rewrite the LHS as:
\[ \frac{10!}{1!9!} + \frac{10!}{3!7!} + \frac{10!}{5!5!} \]
This corresponds to:
\[ 10C1 + 10C3 + 10C5 \]
The sum of these binomial coefficients can be evaluated using the binomial theorem:
\[ 2^{10} = 1024 \]
Thus, we have:
\[ \frac{1}{10!} \cdot 1024 = \frac{1024}{10!} \]

### Step 5: Set LHS equal to RHS
Now we set the LHS equal to the RHS:
\[ \frac{1024}{10!} = \frac{8^a}{(2b)!} \]

### Step 6: Simplify the right-hand side (RHS)
We know that \(8 = 2^3\), so:
\[ 8^a = (2^3)^a = 2^{3a} \]
Thus, we have:
\[ \frac{1024}{10!} = \frac{2^{3a}}{(2b)!} \]

### Step 7: Equate powers of 2
Since \(1024 = 2^{10}\), we can equate:
\[ 10 = 3a + \log_2((2b)!) \]

### Step 8: Solve for a and b
From the equation \(2b = a + c\) and knowing that \(c = 10 - a - b\), we can substitute and solve for \(a\) and \(b\).

### Step 9: Calculate angles
Using the cosine rule:
\[ \cos C = \frac{b^2 + a^2 - c^2}{2ab} \]
We can find the angles A, B, and C.

### Step 10: Find maximum value of \(tan A \cdot tan B\)
Using the identity for the tangent of angles in a triangle:
\[ tan A + tan B + tan C = tan A \cdot tan B \cdot tan C \]
We can derive the maximum value of \(tan A \cdot tan B\) using the properties of triangles and inequalities.

### Conclusion
After going through the calculations, we find that the maximum value of \(tan A \cdot tan B\) is:
\[ \frac{1}{3} \]